动圈与非动圈的JAVA弹性碰撞

JAVA elastic collision of moving and non moving circles(动圈与非动圈的JAVA弹性碰撞)

问题描述

我正在尝试编写一个 Java 移动应用程序 (J2ME)，但遇到了一个问题:在我的项目中，有称为镜头的移动圆圈和称为球体的非移动圆圈.当一个球击中球体时，它应该按照经典物理定律反弹.但是我找不到这种算法.

I'm trying to write a java mobile application (J2ME) and I got stuck with a problem: in my project there are moving circles called shots, and non moving circles called orbs. When a shot hits an orb, it should bounce off by classical physical laws. However I couldn't find any algorithm of this sort.

镜头的运动由 x 轴和 y 轴上的速度(像素/更新)来描述.所有关于圆的信息都是已知的:它们的位置、半径和镜头的速度(在 x 轴和 y 轴上).

The movement of a shot is described by velocity on axis x and y (pixels/update). all the information about the circles is known: their location, radius and the speed (on axis x and y) of the shot.

注意:碰撞后球体不会开始移动，它会停留在原位.碰撞是两者之间的弹性碰撞，而球体保持静止

Note: the orb does not start moving after the collision, it stays at its place. The collision is an elastic collision between the two while the orb remains static

这里是Shot类中的碰撞解决方法:

here is the collision solution method in class Shot:

public void collision(Orb o)
{
    //the orb's center point
    Point oc=new Point(o.getTopLeft().x+o.getWidth()/2,o.getTopLeft().y+o.getWidth()/2);
    //the shot's center point
    Point sc=new Point(topLeft.x+width/2,topLeft.y+width/2);

    //variables vx and vy are the shot's velocity on axis x and y
    if(oc.x==sc.x)
    {
        vy=-vy;
        return ;
    }

    if(oc.y==sc.y)
    {
        vx=-vx;
        return ;
    }

    // o.getWidth() returns the orb's width, width is the shot's width

    double angle=0;  //here should be some sort of calculation of the shot's angle
    setAngle(angle);
}

public void setAngle(double angle)
{
    double v=Math.sqrt(vx*vx+vy*vy);
    vx=Math.cos(Math.toRadians(angle))*v;
    vy=-Math.sin(Math.toRadians(angle))*v;
}

提前感谢所有帮助者

推荐答案

在碰撞点，动量、角动量和能量被保存下来.设m1，m2为圆盘的质量，p1=(p1x,p1y)，p2=(p2x,p2y)为碰撞时圆盘中心的位置，u1,u2为碰撞前的速度，v1,v2为碰撞后的速度碰撞.然后守恒定律要求

At the point of collision, momentum, angular momentum and energy are preserved. Set m1, m2 the masses of the disks, p1=(p1x,p1y), p2=(p2x,p2y) the positions of the centers of the disks at collition time, u1, u2 the velocities before and v1,v2 the velocities after collision. Then the conservation laws demand that

0 = m1*(u1-v1)+m2*(u2-v2)
0 = m1*cross(p1,u1-v1)+m2*cross(p2,u2-v2)
0 = m1*dot(u1-v1,u1+v1)+m2*dot(u2-v2,u2+v2)

使用第一个方程消除 u2-v2

Eliminate u2-v2 using the first equation

0 = m1*cross(p1-p2,u1-v1)
0 = m1*dot(u1-v1,u1+v1-u2-v2)

第一个告诉我们 (u1-v1) 和因此 (u2-v2) 是 (p1-p2) 的倍数，脉冲交换是在法线或径向方向，没有切向相互作用.冲动和能量守恒现在导致相互作用常数a，因此

The first tells us that (u1-v1) and thus (u2-v2) is a multiple of (p1-p2), the impulse exchange is in the normal or radial direction, no tangential interaction. Conservation of impulse and energy now leads to a interaction constant a so that

u1-v1 = m2*a*(p1-p2)
u2-v2 = m1*a*(p2-p1)
0 = dot(m2*a*(p1-p2), 2*u1-m2*a*(p1-p2)-2*u2+m1*a*(p2-p1))

导致非零交互项 a

2 * dot(p1-p2, u1-u2) = (m1+m2) * dot(p1-p2,p1-p2) * a

现在可以使用分数来解决

which can now be solved using the fraction

b = dot(p1-p2, u1-u2) / dot(p1-p2, p1-p2)

作为

a = 2/(m1+m2) * b

v1 = u1 - 2 * m2/(m1+m2) * b * (p1-p2)
v2 = u2 - 2 * m1/(m1+m2) * b * (p2-p1)

要让第二个圆盘静止，设置 u2=0 并且它的质量 m2 非常大或无限大，然后第二个公式说 v2=u2=0 和第一个

To get the second disk stationary, set u2=0 and its mass m2 to be very large or infinite, then the second formula says v2=u2=0 and the first

v1 = u1 - 2 * dot(p1-p2, u1)/dot(p1-p2, p1-p2) * (p1-p2)

<小时>

也就是说，v1 是 u1 在以 (p1-p2) 为法线的平面上的反射.请注意，碰撞点的特征是 norm(p1-p2)=r1+r2 或

dot(p1-p2, p1-p2) = (r1+r2)^2

这样分母就已经从碰撞检测中知道了.

so that the denominator is already known from collision detection.

根据您的代码，oc{x,y} 包含固定磁盘或球体的中心，sc{x,y} 包含中心和 {vx,vy} 移动磁盘的速度.

Per your code, oc{x,y} contains the center of the fixed disk or orb, sc{x,y} the center and {vx,vy} the velocity of the moving disk.

1. 计算 dc={sc.x-oc.x, sc.y-oc.y} 和 dist2=dc.x*dc.x+dc.y*dc.y

1.a 检查 sqrt(dist2) 是否足够接近 sc.radius+oc.radius.普遍的传说说比较正方形更有效.如果 dist2 太小，微调交点的位置.

1.a Check that sqrt(dist2) is sufficiently close to sc.radius+oc.radius. Common lore says that comparing the squares is more efficient. Fine-tune the location of the intersection point if dist2 is too small.

计算 dot = dc.x*vx+dcy*vy 和 dot = dot/dist2

更新vx = vx - 2*dot*dc.x, vy = vy - 2*dot*dc.y

特殊情况包含在这些公式中，例如，对于 dc.y==0，即 oc.y==sc.y 得到 dot=vx/dc.x，所以 vx=-vx, vy=vy 结果.

The special cases are contained inside these formulas, e.g., for dc.y==0, that is, oc.y==sc.y one gets dot=vx/dc.x, so that vx=-vx, vy=vy results.

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